512 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xix 



z/y = ra/(sp), y = H if 



(6) sV+a 2 rsp<r+& 2 rV = D, 



or 6 2 rVH ----- \-ku*=\3, & = b 2 r 2 aa 2 &rs+a 2 s 2 . This quartic is made a 

 square in a special way for special values of m and n for which k D. For 

 the case 61/6 = D=d 2 , make <r = s by taking u = dn, v = m. Then (1) be- 

 comes p 2 +a 2 rp+6 2 r 2 = D = (p re//) 2 , if 2et-\-a 2 t 2 = r}n, e' 2 b 2 t 2 = p/n, which 

 are linear in m and n. 



An anonymous writer 656 gave an elegant solution for the case 

 5 = ^ = 652 = 1. Take x = nR, y = m? n 2 , z = nS, where R = an2m, 

 S = a 2 n + 2m. Then 



Also, /3 = 7i 2 (p 2 +aip<?+# 2 ) 2 if R = p'* q 2 , S = a^-}-2pq. Comparing the two 

 expressions for R and the two for S, we get m and n as fractions whose 

 common denominator is 2(a 2 : Fa), which may be omitted since x, y, z are 

 homogeneous in m and n. For a = a 2 , use the lower sign. 



J. Whitley and W. Rutherford 66 equated p~x-+xy-{-y- and p~z-+xz+x 2 

 to the squares of px+y a and pz+x b, finding x and y in terms of 0. 

 Then p 2 ?/ 2 +?/2+2 2 = D if a quartic in z is a square. 



W. Lenhart 67 took x = abc, y = bdf, z-cfn in (5). By Lagrange's Addi- 

 tion, 90, to Euler's Algebra (Lagrange 63 of Ch. XX), the resulting func- 

 tions are squares if 



df, 2p l q i +ql=fn, 



To solve the equations in the first column, set p-\-q = a, p q = c, 2p+q = d/r, 

 q = rf. From the two values for 2p and the two for q, we get c = a 2rf, 

 d = r(2a rf) . Similarly, by the equations in the second column, b = a 2s/, 

 n = s(2a s/). By the two in the third column, 2p 2 = d-}-b = tc n(t, 

 2q 2 = d b = 2n(t. Eliminating t, we get (3d-f6)(d 6) =4nc. Inserting 

 the earlier values of c, d, b, n, we get 



{(6r+l)a-(2 S +3r 2 )/}{(2r-l)a+(2s-r 2 )/}=8s(a-2r/)(a-is/). 



The final factor will occur also on the left if 2s+3r 2 = |s(6r+l). Then 



a = 12r 3 (5-r) + 5r 2 , /=12r 2 +(3-2r)-2r-l. 



Next, for (4), equate the last two functions to A 2 and B-. Their differences 

 are equal if A-\-B = 2(x-\-y z), A B = ^(x y). Insert the resulting value 

 of B into y*-yz+z* = B z . Thus z = (3x-+Wxy+3y z )/ {8(x +?/)}. Finally, 

 x z xy+y z = D if x = p 2 q 2 , y = 2pq q 2 . 



T. Strong (p. 301) equated (x+y^-Axy, (x+z)--Bxz, (y+z)--Dyz 

 to the squares of x-\-y a, x-\-z b, y -\-z-c. By the first two we get y 

 and z in terms of x. Then the third condition states that two quadratic 

 functions of x are equal. We may equate the constant terms or the coeffi- 

 cients of x 2 and get x rationally. 



66b New Scries of the Math. Repository (ed., T. Leybourn), 3, 1814, I, 151-3. Slightly 



modified solution by A. Martin, The Analyst, Dos Moines, 5, 1S78, 124-5. 

 66 Ladies' Diary, 1834, 37-8, Quest. 1560. 

 87 Math. Miacellany, Flushing, N. Y., 1, 1S3G, 299-301. 



